Probability question

[Cha0s]

is it possible to get P(AUB) from P(A'UB') = 0.2 ?

given is P(A)=0.4 and P(AnB)=0.1

 

[bret]

what the fik?

 

[Cha0s]

anyone? or at least help me find a link with formulas to manipulate this

 

[moshquerade]

the probability of you getting laid is >0

 

[Evadman]

Originally posted by: moshquerade
the probability of you getting laid is >0

I really would have expected an insult there.

 

[helo7050]

????????????????????????????????????????????????????????

 

[SarcasticDwarf]

die in a fire

 

[ribbon13]

OMFG UTF8, get over it...

 

[everman]

It's 42.

 

[bandana163]

Look up the rules in your textbook and solve it.

 

[DaveSimmons]

I'd answer but UTF8 and doing people's homework for them both annoy me.

 

[walla]

Correct me if I'm wrong...

P(A'UB')==1-P(AnB)

Use a venn diagram. Fill in all space that his not A or not B, you will see the only section that is not filled in is where A and B overlap (AnB). That represents 1-P(AnB).

In that case, your problem setup is wrong since you state P(A'UB') != 1-P(AnB).

 

[Cha0s]

Originally posted by: walla
Correct me if I'm wrong...

P(A'UB')==1-P(AnB)

Use a venn diagram. Fill in all space that his not A or not B, you will see the only section that is not filled in is where A and B overlap (AnB). That represents 1-P(AnB).

In that case, your problem setup is wrong since you state P(A'UB') != 1-P(AnB).

(A'UB')= (AnB)'
We have to assume that P(AnB) = 0.1 P(A)=0.4 and P(A'nB')= 0.2

 

[notfred]

Originally posted by: DaveSimmons
I'd answer but UTF8 and doing people's homework for them both annoy me.

Do you even know what UTF-8 is?

 

[walla]

Originally posted by: Cha0s

Originally posted by: walla
Correct me if I'm wrong...

P(A'UB')==1-P(AnB)

Use a venn diagram. Fill in all space that his not A or not B, you will see the only section that is not filled in is where A and B overlap (AnB). That represents 1-P(AnB).

In that case, your problem setup is wrong since you state P(A'UB') != 1-P(AnB).

(A'UB')= (AnB)
We have to assume that P(AnB) = 0.1 P(A)=0.4 and P(A'nB')= 0.2

(A'UB') != (AnB). (for mutually exclusive events)

Think about it...if the following are mutually exclusive
A=Its sunny
B=Its Monday

P(sunny AND {n} Monday) ?= P(cloudy or {U} not Monday)?. No. The two clauses are mutually exclusive. If its cloudy or, for example, Tuesday, it is impossible for it to be also Sunny AND Monday at the same time.

However, taken every single scenario for weather and day of the week, and you take away all of them that are cloudy and then all that are not Monday then all you have left is sunny and Monday.

Hence

P(A'UB')==1-P(AnB)

However, if the events are not mutually exclusive (say for instance A=Sunny, B=Dry), then this formula does not hold since event A effects probability of event B. In that case, you need to know more about conditional probabilities to answer the problem.